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Question Number 29159 by abdo imad last updated on 04/Feb/18
let give f(x)=2(√(x−1)) +3x find f^(−1) (x) and (f^(−1) )^′ (x) .
$${let}\:{give}\:{f}\left({x}\right)=\mathrm{2}\sqrt{{x}−\mathrm{1}}\:+\mathrm{3}{x}\:\:\:{find}\:{f}^{−\mathrm{1}} \left({x}\right)\:{and}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)\:. \\ $$
Commented by abdo imad last updated on 08/Feb/18
f(x)=y ⇔ x=f^(−1) (y) ⇔ 2(√(x−1)) +3x=y ⇔2(√(x−1)) =y−3x ⇔4(x−1)=(y−3x)^2 with the cond. x−1≥0 andy−3x≥0 ⇔ 4x−4=y^2 −6xy +9x^2 ⇔ 9x^2 −6xy −4x +y^2 +4=0 ⇔ 9x^2 −2(3y+2)x +y^2 +4=0 Δ^′ = (3y+2)^2 −9(y^2 +4)=9y^2 +12y +4 −9y^2 −36 =12y −32 =4(3y −8) the equ.have roots⇔ y≥(8/3) in this case x_1 =((3y+2 +2(√(3y−8)))/9) and x_2 =((3y+2 −2(√(3y−8)))/9) we must have 3x−y≤0 and x≥1 3x_1 −y=((3y +2 +2(√(3y−8)) −3y)/3) =((2+2(√(3y−8)))/3) >0 (x_1 to eliminate) so f^(−1) (x)=(1/9)(3x+2 −2(√(3x−8)) ) and for x>(3/8) (f^(−1) )^′ (x)=(1/9)(3 −2 (3/(2(√(3x−8))))) ⇒ (f^(−1) )^′ (x)=(1/3)( 1− (1/( (√(3x−8))))) .
$${f}\left({x}\right)={y}\:\Leftrightarrow\:{x}={f}^{−\mathrm{1}} \left({y}\right)\:\Leftrightarrow\:\mathrm{2}\sqrt{{x}−\mathrm{1}}\:+\mathrm{3}{x}={y} \\ $$$$\Leftrightarrow\mathrm{2}\sqrt{{x}−\mathrm{1}}\:={y}−\mathrm{3}{x}\:\Leftrightarrow\mathrm{4}\left({x}−\mathrm{1}\right)=\left({y}−\mathrm{3}{x}\right)^{\mathrm{2}} \:{with}\:{the}\:{cond}. \\ $$$${x}−\mathrm{1}\geqslant\mathrm{0}\:{andy}−\mathrm{3}{x}\geqslant\mathrm{0}\:\Leftrightarrow\:\mathrm{4}{x}−\mathrm{4}={y}^{\mathrm{2}} \:−\mathrm{6}{xy}\:+\mathrm{9}{x}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{9}{x}^{\mathrm{2}} \:−\mathrm{6}{xy}\:−\mathrm{4}{x}\:+{y}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{9}{x}^{\mathrm{2}} \:−\mathrm{2}\left(\mathrm{3}{y}+\mathrm{2}\right){x}\:+{y}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\Delta^{'} =\:\left(\mathrm{3}{y}+\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{9}\left({y}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{9}{y}^{\mathrm{2}} +\mathrm{12}{y}\:+\mathrm{4}\:−\mathrm{9}{y}^{\mathrm{2}} −\mathrm{36} \\ $$$$=\mathrm{12}{y}\:−\mathrm{32}\:=\mathrm{4}\left(\mathrm{3}{y}\:−\mathrm{8}\right)\:{the}\:{equ}.{have}\:{roots}\Leftrightarrow\:{y}\geqslant\frac{\mathrm{8}}{\mathrm{3}}\:{in}\:{this} \\ $$$${case}\:\:{x}_{\mathrm{1}} =\frac{\mathrm{3}{y}+\mathrm{2}\:+\mathrm{2}\sqrt{\mathrm{3}{y}−\mathrm{8}}}{\mathrm{9}}\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{3}{y}+\mathrm{2}\:−\mathrm{2}\sqrt{\mathrm{3}{y}−\mathrm{8}}}{\mathrm{9}} \\ $$$${we}\:{must}\:{have}\:\mathrm{3}{x}−{y}\leqslant\mathrm{0}\:{and}\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{3}{x}_{\mathrm{1}} \:−{y}=\frac{\mathrm{3}{y}\:+\mathrm{2}\:+\mathrm{2}\sqrt{\mathrm{3}{y}−\mathrm{8}}\:−\mathrm{3}{y}}{\mathrm{3}}\:=\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}{y}−\mathrm{8}}}{\mathrm{3}}\:>\mathrm{0}\:\left({x}_{\mathrm{1}} {to}\right. \\ $$$$\left.{eliminate}\right)\:{so}\:{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{3}{x}+\mathrm{2}\:−\mathrm{2}\sqrt{\mathrm{3}{x}−\mathrm{8}}\:\right)\:{and}\:{for} \\ $$$${x}>\frac{\mathrm{3}}{\mathrm{8}}\:\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{3}\:−\mathrm{2}\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{3}{x}−\mathrm{8}}}\right)\:\Rightarrow \\ $$$$\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\:\mathrm{1}−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}{x}−\mathrm{8}}}\right)\:. \\ $$

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